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Hydro turbine water input/output

  1. Jan 8, 2012 #1
    having a problem here: If a hydro turbine is rated 1 MW @ q=5m3/s with 25 m total head does that mean that the water once through the turbine will also be q=5m3/s? Lets assume that we have ideal maximum conditions including the most efficient nozzles. I have seen some examples where the flow past the turbine is much less than the water in. Your comments would greatly be appreciated.
     
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  3. Jan 8, 2012 #2

    russ_watters

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    Water in has to equal water out.
     
  4. Jan 8, 2012 #3
    If the tubine is submerged and the end of the draft tube is 70' below the surface how much is the decrease of Water in/water out?
     
  5. Jan 8, 2012 #4

    boneh3ad

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    Unless you are implying that the turbine is somehow destroying matter or storing water somewhere, then the water in has to equal the water out. There is no way around that.
     
  6. Jan 8, 2012 #5
    OK; I understand; so lets rephrase the above question: If the turbine is submerged and the end of the draft tube is 70' below the surface the water intake/output is the same. How much does the depth of operation reduce intake output? For example: under normal (traditional) conditions 5m3/s goes in and 5m3/s goes out. Now under 70 feet of water how does this change? Is it now 2m3/s in and 2 ms3/s out? Or is it the same 5m3/s in and 5m3/s out?
     
  7. Jan 9, 2012 #6

    russ_watters

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    Again, flow rate must be the same, but I rather suspect the real issue here has to do with the pressure...

    I think there is more here than you are telling us: please elaborate on what your issue is/what you are thinking/trying to do.
     
  8. Jan 9, 2012 #7
    We've established that flow rate must be the same for input and output.I'm simply trying to discharge a hydro turbine 70 feet deep below the surface of the water. Will the rated flow of 5m3/s be reduced in this situation because of the pressure?
     
  9. Jan 9, 2012 #8

    boneh3ad

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    It shouldn't. All that will matter is the pressure differential, and that won't change. Only the ambient pressure changes.
     
  10. Jan 9, 2012 #9
    Thanks, I really appreciate your help.
     
  11. Jan 9, 2012 #10

    russ_watters

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    How do you discharge 70 feet underwater without losing 90% of your pressure differential and thus 90% of your power?

    Whether the turbine will allow the same flow through when you discharge underwater depends on if you still have the same rated delta-p and if it was rated with a free or submerged outlet.

    Is the delta-p still 25m? You aren't trying to turn 3m of head into 25m of head by submerging the outlet 22m, are you?

    Or let me ask the other way: what benefit are you looking to get by discharging so far underwater?
     
    Last edited: Jan 9, 2012
  12. Jan 9, 2012 #11

    boneh3ad

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    Seemed to me that he was just looking at the effect of pressure where the ambient pressure would be ramped up dramatically if the turbine were submerged. For a real turbine, it is really unclear how exactly the turbine would function differently without knowing a few things about it first, but in general, a turbine shouldn't be affected a whole lot by ambient pressure for an incompressible fluid such as water.
     
  13. Jan 9, 2012 #12
    I'm working on some calculations now and will share them with you as early as tomorrow. You see why this is a big question mark? Already One says it won't work very well at all, another says it could or might work OK. What information about the turbine would make this issue clearer? In the meantime please hang on and assist me in making this issue more understandable.
     
  14. Jan 10, 2012 #13

    russ_watters

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    It isn't a big question mark of an engineering problem, it's a pretty simple one. Bonehead and I both understand the issue well enough to know that you haven't provided a complete description of your setup and if you provided it we'd be able to give you a better answer. The difference between our answers is that we're speculating in different directions, but I'm sure we both agree that submerging your outlet neither helps nor hurts your head situation. So what we need is more detail - or better yet, a diagram - of what you are proposing to do.

    Still, I'll bet him a quarter that you're trying to defeat conservation of energy, thinking that submerging the outlet provides you extra head across the turbine. My spidey sense is tingling.

    [edit] Really, we're only missing one piece of information:the location (height) of the inlet.
     
    Last edited: Jan 10, 2012
  15. Jan 10, 2012 #14
    Please consider this drawing to clarify the problem. What else do you need?
     

    Attached Files:

  16. Jan 10, 2012 #15
    By the way, there are new treatments available for chronic acute tingling spidy sense.
     
  17. Jan 10, 2012 #16

    russ_watters

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    Ok, so it is exactly what I said in post #10:

    5-70+70= 5 feet of head, not 75 feet of head.

    If what you were suggesting were possible, you could go a step further and create energy from nothing in any still pond by submerging a a pipe in it completely. Water would just spontaneously flow through the pipe. But see, if the bottom of such a pipe were 70 feet down there'd be 70' of head inside and 70' of head outside, with no extra delta-P available to push water through the pipe. Sorry - this is a pretty common mistake.
     
    Last edited: Jan 10, 2012
  18. Jan 11, 2012 #17
    Please examine the attached ; let me know If this is going in the right direction; Thanks for your excellent help./
     

    Attached Files:

  19. Jan 12, 2012 #18
    OK gentlemen, Thanks for your input. Please look at the attached doc. let me know if it agrees with your thoughts on this subject.
     

    Attached Files:

    • 1MW.doc
      1MW.doc
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  20. Jan 12, 2012 #19
    Are you OK with this calculation?

    Given:
    q1 = 5 m3/s
    h1 = 75 ft = 22.860 m
    h2 = 70 ft = 21.336 m
    = 1000 kg/m3 is density of water
    g = 9.81 m/s2 is acceleration due to gravity

    Solution:
    (1) In absence of water in the tank A the power output of the turbine is equal to decrease in gravitational potential energy of the water which is given by
    W1 = *g*h1*q1 = 1000*9.81*22.86*5 = 1.121*106 W = 1.121 MW

    The speed of water flow is given by
    v1 = = sqr(2*9.81*22.860) = 21.18 m/s

    Let A be cross sectional are of the tube. Then the volume water flow is given by
    q1 = A*v1

    From which we get
    A = q1/v1 = 5/21.18 = 0.2361 m2

    Now consider when the draft tube is at 70 feet below the water surface

    Now the speed of water flow is given by
    v2 = = sqr[2*9.81*(22.860-21.336)] = 5.468 m/s

    So the new volume water flow is given by
    q2 = A*v2 = 0.2361*5.468 = 1.291 m3/s

    Comparing it with q1 we conclude that the volume water flow through the turbine reduces

    (2) The water is incompressible, so the water flow rate is the same in the draft tube. Therefore the output at the end of the draft tube is the same as at its input
    qout = q2 = 1.291 m3/s


    (3) The new output power of the turbine at the end of the draft tube is given by
    W2 = *g*(h1 - h2)*q2 = 1000*9.81*(22.860-21.336)*1.291 = 1.930*104 W = 19.30 kW

    Comparing it with W1 we conclude that the power of the turbine reduces
     
  21. Jan 14, 2012 #20

    russ_watters

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    It looks like you are calculating the velocity of the steam by assuming all of the potential energy is converted to kinetic energy, then assuming your pipe is sized for that velocity and flow rate and calculating its size. These assumptions are improper. Even if the turbine is a velocity type (Pelton), the only place the gravitational potential energy is all converted to kinetic energy is in the turbine itself, at the nozzles. Everywhere else in the system, the pipes would be sized much, much larger to avoid excessive pressure losses due to the high velocity.

    However, for a Pelton turbine with a fixed nozzle size of .2361 m^2 and no losses, I believe it to be a reasonable calculation.

    However, however, for a real turbine with losses or for a normal turbine which relies more on static pressure than velocity pressure, you'll really want to look at the performance curves to know what the new performance will be. Real performance can vary greatly from the theoretical.
     
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